Question

a) How many mL of 0.1 M Sr(OH)2 will be added to a beaker containing 150 mL of 0.16 M HClO4 (aq) to reach the equivalence point?

b) What is the pH of a beaker containing 150 mL of 0.16 M solution of CH3COOH (aq) Ka=1.7*10^-5 titrated with 0.2 M NaOH at the equivalence point?

c) Would you expect the volume to reach the equivalence point to be the same or different for 150 mL of 0.16 M HClO4 being titrated with 0.1 M Sr(OH)2 and 150 mL 0.16 M solution of CH3COOH titrated with 0.2 M NaOh? Why?

Answer 1

a) How many mL of 0.1 M Sr(OH)2 will be added to a beaker containing 150 mL of 0.16 M HClO4 (aq) to reach the equivalence point?

mol of acid = MV = 150*0.16 = 24mmol of acid

ratio si 1:2 so we need half of Sr(OH)2

then

24/2 = 12 mmol of base

M = mmol/ mL

mL = mmol/M = 12/0.1 = 120 ml of base

b) What is the pH of a beaker containing 150 mL of 0.16 M solution of CH3COOH (aq) Ka=1.7*10^-5 titrated with 0.2 M NaOH at the equivalence point?

mol of acid = MV = 150*0.16 = 24 mmol of acid

mol of base = MV = 0.2*V = 24

V = 24/0.2 = 120ml of base

total V = 120 + 150 = 270 ml

then

CH3COO- + H2O = CH3COOH +OH-

Kb = [H+][OH-]/[CH3COO]

Kb = (10^-14)/(1.7*10^-5) =5.88235*10^-10

then

5.88235*10^-10 = x*x/(0.16/2 - x)

x = 6.85*10^-6

pOH =-log(x) =-log(6.85*10^-6) = 5.164

ph = 14-5.164= 8.836

c)

c) Would you expect the volume to reach the equivalence point to be the same or different for 150 mL of 0.16 M HClO4 being titrated with 0.1 M Sr(OH)2 and 150 mL 0.16 M solution of CH3COOH titrated with 0.2 M NaOh? Why?

Same Volume used, different pH!